Question

Find the exact length of the curve. x = 1 + 6t2, y = 4 + 4t3, 0 ≤ t ≤ 3

Answer 1

$x=1+6t^2\implies\dfrac{\mathrm dx}{\mathrm dt}=12t$

$y=4+4t^3\implies\dfrac{\mathrm dy}{\mathrm dt}=12t^2$

The length of the curve $C$ is given by

$\displaystyle\int_C\mathrm ds=\int_0^3\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

$=\displaystyle\int_0^3\sqrt{144t^2+144t^4}\,\mathrm dt$

$=\displaystyle12\int_0^3t\sqrt{1+t^2}\,\mathrm dt$

$=\displaystyle6\int_0^3 2t\sqrt{1+t^2}\,\mathrm dt$

$=\displaystyle6\int_0^3\sqrt{1+t^2}\,\mathrm d(1+t^2)$

$=6\cdot\dfrac23(1+t^2)^{3/2}\bigg|_0^3$

$=4(10^{3/2}-1)$

$=40\sqrt{10}-4$